A prism has a refracting angle of 30^circ and | vedclass

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Question

(D) Since the incident ray retraces its path,it must strike the silvered face of the prism normally (at an angle of $90^\circ$).In the triangle formed

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Since the incident ray retraces its path,it must strike the silvered face of the prism normally (at an angle of $90^\circ$).In the triangle formed by the prism,let the vertices be $A, B, C$ and the point of incidence be $D$ on face $AC$,and the point on the silvered face be $E$. In $	riangle AED$,the angles are $30^\circ$,$90^\circ$,and $\angle ADE$.Sum of angles in $	riangle AED = 180^\circ$.$30^\circ + 90^\circ + \angle ADE = 180^\circ \Rightarrow \angle ADE = 60^\circ$.The angle of refraction $r$ is the angle between the normal at $D$ and the refracted ray $DE$. Since the normal is perpendicular to the face $AC$,we have $\angle r + \angle ADE = 90^\circ$.$\angle r = 90^\circ - 60^\circ = 30^\circ$.Applying Snell's Law at the surface $AC$:$1 \cdot \sin i = \mu \cdot \sin r$$\sin i = \sqrt{2} \cdot \sin 30^\circ$$\sin i = \sqrt{2} \cdot \frac{1}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = 45^\circ$.

$A$ prism has a refracting angle of $30^\circ$ and a refractive index of $\mu = \sqrt{2}$. One of its faces is silvered. $A$ ray of light incident on the other face retraces its path. What is the angle of incidence in degrees?

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